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Chapter 5

Applying Newton’s Laws

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Newton’s Laws

2nd Law:

1st Law: An object at rest or traveling in uniform motion will remain at rest or traveling in uniform motion unless and until an external force is applied

amFnetrr

=

zznetyynetxxnet maFmaFmaF === ,,, ,,

3rd Law: For every Action, there is an equal but opposite Reaction

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Part 1

Some particular forces

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Some particular forcesIn Chapter 5 we deal with following forces

Gravitational force FgNormal force NTension TFrictional force fCentripetal force Fc

new concepts

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5

The Gravitational ForceA gravitational force on a body is a pull that is directed toward a second body.In chapter 5 we do not discuss the nature of this force, and we usually consider that the second body is EarthWe mean that the gravitational force pulls directly toward the center of Earth – that is, directly down toward the ground

new concepts

mgFg =

ground

Magnitude:

Direction: toward the ground

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The Normal ForceWhen a body presses against a surface, the surface deforms and pushes on the body with a normal force that is perpendicular to the surface

new concepts

Magnitude: comp. of mg ⊥ to the surface

Direction: perpendicular to the surface

mgN =

gFr

Nr

gFr

Nr

α

)cos(αmgN =

4

7

Magnitude:

Direction: away from the body and along the cord

Tension forceWhen a cord (a rope, cable, …) is attached to a body a pulled taut, the cord pulls on the body with force T. Normally: the cord is considered massless (comparing to the body’s mass) and unstretchable (it si only a connection between bodies)

new concepts

T

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Part 1b

Frictional forces

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9

Frictional ForcesJust one exampleThe bad:About 20% of the gasoline used in an automobile is needed to counteract friction in the engine and in the drive train.Then … air resistanceThe goodWithout frictional forces we could not get an automobile to go anywhere

new concepts

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Frictional Forces (cont)Microscopically, friction is a very complicated phenomenon(spikes, cold welding, …)Macroscopically, at large-scale level, it is relatively simplePhenomenologically (empirically), force of friction acting on a body is directly proportional in magnitude to normal force of the surface on the body:

Nfrr

∝

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11

Property 1: If the body does not move, then the static frictional force and the component of that is parallel to the surface balance each other.

Property 2: The magnitude of has a maximum value fs,max that is given by

where is a coefficient of static friction, and N is the magnitude of the normal force on the body from the surface

Force of static friction: one must overcome (exceed) it in order to initiate motion of the body along the surface

Properties of Friction

sfr

Fr

sfr

Nf ss μ=max,

sμ

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Formal derivationWhen block is not moving, friction force compensates x-component of gravitational force: fs = mg*sin(θ) . However by definition fs=μsN = μsmg*cos(θ). Then mg*sin(θ) = μsmg*cos(θ), and μ = tan(θ)

gmr

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13

Static FrictionNf ss μ=max,

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Property 3: If the body begins to slide along the surface, the magnitude of the frictional force rapidly decreases to a value fk given by

where is the coefficient of kinetic friction

Force of kinetic friction: acts while the body moves along the surface

Properties of Friction (more)

Nf kk μ=

kμ

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Static and Kinetic Friction

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Frictional Forces

Static Friction

Direction: parallel to the surface, and is directed opposite the component of an external force.Magnitude

Kinetic Friction

Direction: always opposite to direction of velocityMagnitude

Nf ss μ=

Nf kk μ=

Frictional force is independent of the area of contact between the body and the surface.

summary

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A fluid is anything that can flow – either a gas or liquid

When a body moves through the fluid (or the fluid moves past the body), the body experience a drag force that opposes the relative motion.

At low speed:

At high speed

Terminal speed: when a body falls at constant speed, i.e.

How about motion in fluids

vkf 1=

222 5.0 AvCDvkf ρ===

ACmgv

mgAvC

ρ

ρ

2

5.0 2

=

= C – the drag coefficient (0.2 – 1.0)A – effective cross sectionρ - the air density

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Terminal speed

3115parachutist

6157raindrop

474420basketball

2109242baseball

43013060sky diver

2500316145shot

distance (m) 95%

speed(mph)

speed(m/s)object

ACmgvρ

2=

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Part 2

Free-Body Diagrams

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Free-body diagram as a powerful toolIDEA: replace an actual environment of an object as a set of forces acting on that object

new concepts

gFr

Nr

leftFr

rightFr

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Free-body diagram as a powerful toolFree-body diagrams show all forces acting on a body

Key ideas for drawing a free-body diagram:1. Include: ALL forces acting on the body matter.2. When a problem includes more than one body – draw a separate free-body diagram for each body.3. Not to include: any forces that the body exerts on any other body.4. Not to include: non-existing forces (no object – no force).

new concepts

∑=++=n

iin FFFFFrr

Krrr

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examplea coin on a book

example

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Part 3

Problems with gravitational, normal and tension forces

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Objects in equilibriumAccording to Newton’s first law for an object in equilibrium (at rest)

0

0

0

21

21

21

==++=

==++=

==++=

∑

∑

∑

n

iyiynyyy

n

ixixnxxx

n

iin

FFFFF

FFFFF

FFFFF

K

K

rrK

rrr

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Objects in motionAccording to Newton’s second law

y

n

iyiynyyy

x

n

ixixnxxx

n

iin

maFFFFF

maFFFFF

amFFFFF

==++=

==++=

==++=

∑

∑

∑

K

K

rrrK

rrr

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21

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example 1: one object in 1Dan object on a string (equilibrium)Given: mUnknown: T

example

m

mg

T0=+= gmTFnetrrr

mgT

mgTmama

y

x

=⎩⎨⎧

=−==

00

14

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example 2: two objects in 1Dtwo objects on a string (equilibrium)Given: m1, m2Unknown: T1, T2

example

m2

m1

m2g

m1g

T2

T1

0

0

222,

1211,

=+=

=++=

gmTF

gmTTF

net

netrrr

rrrr

00

22

121

=−=−−

gmTgmTT

gmTgmmT

22

211 )(=

+=

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example 3a: 2D caseone object in equilibriumGiven: m, angleUnknown: T1, T2, T3Objects: the engine & the ring

example

0)sin(:0)cos(:

0:

132

232

11

=−=−

=−

TTyTTx

mgTy

αα

0

0

3212,

11,

=++=

=+=

TTTF

gmTF

net

netrrrr

rrr

mg

)sin(/)cos()sin(/

2

3

1

ααα

mgTmgTmgT

===

x

y

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example 3b: 2D caseGiven: m, anglesUnknown: T, T2

example

mg

T1

T2

T1

T3

α

0

0

322,

11,

=++=

=+=

TTTF

gmTF

net

netrrrr

rrr

0)sin()sin(:0)cos()cos(:

0:

1322

322

11

=−+=−

=−

TTTyTTx

mgTy

αααα

)sin(232

1

αmgTT

mgT

==

= )sin(2/1 αα300 1.00100 2.88

50 5.7410 28.6 practical applications?

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example 4a: one object in 2Dobject in equilibriumGiven: m, angleUnknown: T, N

example

0=++= gmTNFnetrrrr

0)cos(:0)sin(:=−=−

θθ

mgNymgTx

)cos()sin(θθ

mgNmgT

==

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example 4b: one object in 2DMoving object (no friction)Given: m, angleUnknown: a

example

amgmNFnetrrrr

=+=

0)cos(:)sin(:

==−=

y

x

mamgNymamgxθ

θ

)cos()sin(θ

θmgNgax

==

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example 5: two objects in 2DMoving objects (no friction)Given: M, mUnknown: T, a

example

aMTF

amgmTF

Mnet

mnetrrr

rrrr

==

=+=

,

,

mamgTyMaTx

−=−=

::

gMm

mMT

gMm

ma

+=

+=

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example 6: three objects in 1DMoving objects (no friction)Given: T3, m1, m2, m3Unknown: a, T1, T2

Attention: all blocks have the same acceleration

example

amTTamTT

amT

323

212

11

=−=−

=

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Part 4

Problems with gravitational, normal, tension forces and frictional forces

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example: one object in 1DThe record for the longest skid marks on a public road was recorded in 1960 by a Jaguar on the M1 highway in England – the marks were 290 meters long.Given: L = 290 m, kinetic friction 0.6Unknown: v0.Assume: the car’s deceleration was constant during the breaking

example

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example: one object in 1DGiven: L = 290 m, kinetic friction 0.6Unknown: v0.

example

)(2

)(2

00

020

2

xxgv

gamgfma

xxavv

k

k

kk

−=

−===−

−+=

μ

μμ

v0 = 58 m/s = 130 mph The result does NOT depend on the car’s mass!!!

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example: a book against a wallGiven: book 2 kg, static friction 0.4Unknown: force to keep the book

example

s

s

s

mgFFNf

fmg

μμμ

/===

=

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Part 5

Dynamics of Uniform Circular Motion

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example: “Dare Devil 1901”In 1901 circus performance Given: R, MUnknown: v

example

amgmN rrr

=+

gRvN

mgRvmN

RvmmgN

==

−=

−=−−

:0

2

2

for R = 2.7 m v = 5.1 m/s

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